TITLE: Enlightening proof that the algebraic numbers form a field
QUESTION [43 upvotes]: The proof I'm familiar with that the algebraic numbers $\mathbb A$ form a field uses the fact that the resultant of two polynomials $p,q\in\mathbb Q[x]$ satisfies the following properties:

It is $0$ iff $p$ and $q$ have a common factor.
It is a polynomial in the coefficients of $p$ and $q$.

We then introduce a new variable and cleverly manipulate $p$ and $q$ to get polynomials which vanish at the sums and products of their roots. This is in some ways a nice proof, e.g. it is constructive and so can be converted into an algorithm to find such polynomials (which I in fact just finished doing in C). But I don't find it very enlightening; it seems like the fact that $\mathbb A$ is a field is simply an accident. Is there a more enlightening proof of this fact?

REPLY [18 votes]: I find another charming proof in Lecture on the theory of algebraic numbers by Hecke. It is constructive in appearance, but is impractical in fact.
Let $p(x)$ and $q(x)$ be two polynomials with rational coefficients, and with roots $\alpha$ and $\beta$ respectively. Let $\{\alpha_i|i=0,\ldots,m\}$ and $\{\beta_j|j=0,\ldots,n\}$ be their conjugates, i.e. roots of $p$ and $q$ respectively. Then set $$r(x)=\prod_{i=0}^m\prod_{j=0}^n(x-\alpha_i-\beta_j).$$ We kow that $\alpha+\beta$ is a root of $r(x)$, and that coefficients of $r(x)$ are symmetric functions of roots of $p$ and $q$, and hence could be written as polynomials of coefficients of the two polynomials, thus is rational. Therefore $\alpha+\beta$ is also algebraic. A similar proof goes for products. Finally, write $$p(x)=\sum_0^ma_ix^i.$$Then $$\sum_0^ma_i\alpha^i=0,$$ so that $$\sum_0^ma_{m-i}\alpha^{-i}=0,$$ and hence $\alpha^{-1}$ also is algebraic.
Inform me of any ambiguity or error, thanks.